Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $r = \dfrac{z + 6}{5z^2 + 45z + 90} \times \dfrac{-5z^2 - 40z - 75}{z - 8} $
Answer: First factor out any common factors. $r = \dfrac{z + 6}{5(z^2 + 9z + 18)} \times \dfrac{-5(z^2 + 8z + 15)}{z - 8} $ Then factor the quadratic expressions. $r = \dfrac {z + 6} {5(z + 3)(z + 6)} \times \dfrac {-5(z + 3)(z + 5)} {z - 8} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {(z + 6) \times -5(z + 3)(z + 5) } { 5(z + 3)(z + 6) \times (z - 8)} $ $r = \dfrac {-5(z + 3)(z + 5)(z + 6)} {5(z + 3)(z + 6)(z - 8)} $ Notice that $(z + 3)$ and $(z + 6)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {-5\cancel{(z + 3)}(z + 5)(z + 6)} {5\cancel{(z + 3)}(z + 6)(z - 8)} $ We are dividing by $z + 3$ , so $z + 3 \neq 0$ Therefore, $z \neq -3$ $r = \dfrac {-5\cancel{(z + 3)}(z + 5)\cancel{(z + 6)}} {5\cancel{(z + 3)}\cancel{(z + 6)}(z - 8)} $ We are dividing by $z + 6$ , so $z + 6 \neq 0$ Therefore, $z \neq -6$ $r = \dfrac {-5(z + 5)} {5(z - 8)} $ $ r = \dfrac{-(z + 5)}{z - 8}; z \neq -3; z \neq -6 $